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HAL Id: hal-01652890

https://hal.archives-ouvertes.fr/hal-01652890v2

Preprint submitted on 5 Dec 2019

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Geometric and probabilistic results for the observability of the wave equation

Emmanuel Humbert, Yannick Privat, Emmanuel Trélat

To cite this version:

Emmanuel Humbert, Yannick Privat, Emmanuel Trélat. Geometric and probabilistic results for the

observability of the wave equation. 2019. �hal-01652890v2�

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Geometric and probabilistic results for the observability of the wave equation

Emmanuel Humbert

Yannick Privat

Emmanuel Tr´ elat

Abstract

Given any measurable subset ω of a closed Riemannian manifold (M, g) and given any T >0, we define`T(ω)∈[0,1] as the smallest average time over [0, T] spent by all geodesic rays inω. This quantity appears naturally when studying observability properties for the wave equation onM, withωas an observation subset: the condition`T(ω)>0 is the well known Geometric Control Condition.

In this article we establish two properties of the functional`T, one is geometric and the other is probabilistic.

The first geometric property is on the maximal discrepancy of`T when taking the closure.

We may have`T(˚ω)< `T(ω) whenever there exist rays grazingωand the discrepancy between both quantities may be equal to 1 for some subsetsω. We prove that, if the metricgis C2 and ifωsatisfies a slight regularity assumption, then `T(ω)6 12 `T(˚ω) + 1

. We also show that our assumptions are essentially sharp; in particular, surprisingly the result is wrong if the metric g is not C2. As a consequence, ifω is regular enough and if `T(ω) > 1/2 then the Geometric Control Condition is satisfied and thus the wave equation is observable onωin timeT.

The second property is of probabilistic nature. We takeM=T2, the flat two-dimensional torus, and we consider a regular grid on it, a regular checkerboard made ofn2 square white cells. We construct random subsetsωnε by darkening each cell in this grid with a probability ε. We prove that the random law `Tnε) converges in probability to ε as n →+∞. As a consequence, if n is large enough then the Geometric Control Condition is satisfied almost surely and thus the wave equation is observable onωnε in timeT.

1 Introduction and main results

Let (M, g) be a closed connected Riemannian manifold. We denote by Γ the set of geodesic rays, that is, the set of projections ontoM of Riemannian geodesic curves in the co-sphere bundleSM. Given anyT >0 and any Lebesgue measurable subsetω ofM, we define

`T(ω) = inf

γ∈Γ

1 T

Z T 0

χω(γ(t))dt. (1)

Here, χω is the characteristic function of ω, defined by χω(x) = 1 if x ∈ ω and χω(x) = 0 if x∈M \ω. The real number `T(ω) is the smallest average time over [0, T] spent by all geodesic

Institut Denis Poisson, UFR Sciences et Technologie, Facult´e Fran¸cois Rabelais, Parc de Grandmont, 37200 Tours, France (emmanuel.humbert@lmpt.univ-tours.fr).

IRMA, Universit´e de Strasbourg, CNRS UMR 7501, 7 rue Ren´e Descartes, 67084 Strasbourg, France (yannick.privat@unistra.fr).

Sorbonne Universit´e, CNRS, Universit´e de Paris, Inria, Laboratoire Jacques-Louis Lions (LJLL), F-75005 Paris, France (emmanuel.trelat@sorbonne-universite.fr).

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rays inω. The condition

`T(ω)>0

means that all geodesic rays, propagating in M, meetω within time T. This condition, usually calledGeometric Control Condition (in short, GCC), is related to observability properties for the wave equation

tty− 4gy= 0 in (0, T)×M (2)

where4g is the Laplace-Beltrami operator onM for the metricg.

More precisely, denoting bydxgthe canonical Riemannian volume, we define the observability constantCT(ω)>0 as the largest possible nonnegative constant Csuch that the inequality

Z T 0

Z

ω

|y(t, x)|2dxgdt>C

ky(0,·)k2L2(M)+k∂ty(0, x)k2H−1(M)

(3) is satisfied for any solutionyof (2), that is,

CT(ω) = inf (Z T

0

Z

ω

|y(t, x)|2dxgdt | k(y(0,·), ∂ty(0,·))kL2(M)×H−1(M)= 1 )

.

When CT(ω) > 0, the wave equation (2) is said to be observable on ω in time T, and when CT(ω) = 0 we say that observability does not hold for (ω, T).

It has been proved in [1,9] that, forωopen, observability holds if the pair (ω, T) satisfies GCC, i.e., if`T(ω)>0. In other words, ifω is open and satisfies`T(ω)>0 then the wave equation (2) is observable onω in timeT.

The converse is not true: GCC is not a necessary condition for observability. It is shown in [8]

that, ifM =S2 (the unit sphere in IR3 endowed with the restriction of the Euclidean structure), if ω is the open Northern hemisphere, then `T(ω) = 0 for every T > 0, and however one has CT(ω)>0 for every T > π. The latter fact is established by an explicit computation exploiting symmetries of solutions. This failure of the functional `T to capture the observability property is due to the existence, here, of a very particular geodesic ray which isgrazing the open set ω, namely, the equator. In this example, consideering the closureω ofω, it is interesting to observe that`T(ω) = 0 for everyT 6π(take a geodesic ray contained in the closed Southern hemisphere) and`T(ω)>0 for everyT > π, with`T(ω) = 12 whenT >2π. The latter equality is in contrast with`T(ω) = 0: there is thus a discrepancy 1/2 in the value of `T for T >2π when taking the closure ofω. In this specific case, this discrepancy is caused by the equator, which is a geodesic ray grazing the open subsetω.

Our first main result below shows that 1/2 is actually the maximal possible discrepancy.

1.1 A geometric result on the maximal discrepancy of `

T

In general, one can always find subsetsω for which the difference`T(ω)−`T(˚ω) is arbitrary close to 1. Surprisingly, under slight regularity assymptions, this maximal discrepancy is 1/2 only.

Theorem 1. LetT >0be arbitrary and letω be a measurable subset ofM. We make the following assumptions:

(i) The metricg is at least of classC2.

(ii) ω is an embeddedC1submanifold ofM with boundary if n>3and is piecewise C1 ifn= 2.

Then

`T(ω)61

2 `T(˚ω) + 1

. (4)

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We give more details and a number of comments on this theorem in Section2.

At the opposite, as an obvious remark, if there is no geodesic ray grazingω (more precisely, if there is no geodesic ray having a contact of infinite order with∂ω=ω\˚ω) then`T(ω) =`T(˚ω).

In more general, the existence of suchgrazing rays, which are rays having a contact of infinite order with∂ω=ω\˚ω and may involve an arc entirely contained in∂ω, adds a serious difficulty to the analysis of observability (see [1]).

It is noticeable that, if one replaces the characteristic functionχωofωby a continuous function a, in the integral at the left-hand side of (3) (i.e., RT

0

R

Ma(x)|y(t, x)|2dxgdt) as well as in the definition (1) of the functional `T, this difficulty disappears and the condition `T(a)>0 becomes a necessary and sufficient condition for observability of (2) onω in timeT (see [4]).

By the way, for completeness, we provide in Appendix A some semi-continuity properties of the functional`T, which may be of interest for other purposes.

The issue of the observability on a general measurable subsetω⊂M has remained widely open for a long time. Recent advances have been made, which we can summarize as follows. It has been established in [7] that observability on a measurable subsetω in timeT is satisfied if and only if αT(ω)>0. The quantityαT(ω), defined in [7] as the limit of highfrequency observability constants, is however not easy to compute and we have, in general, the inequality`T(˚ω)6αT(ω)6`T(ω). In particular, the condition`T(ω)>0 becomes a necessary and sufficient condition for observability as soon as there are no geodesic rays grazingω. It has also been shown in [7] that limT→+∞CTT(ω) is the minimum of two quantities, one of them being`T(ω) and the other being of a spectral nature.

We have the following corollary of Theorem1, using the fact that, since ˚ωis open, the condition

`T(˚ω)>0 implies observability for (˚ω, T), and thusCT(ω)>CT(˚ω)>0.

Corollary 1. Under the assumptions of Theorem 1, if `T(ω)>1/2 then`T(˚ω)>0 and thus the wave equation (2)is observable on ω in time T.

Note that Corollary 1 does not apply to the (limit) case where M = S2 and ω is the open Northern hemisphere. It does neither apply to the case where M is the two-dimensional torus andω is a half-covering open checkerboard on it, as in [3, 5] (see next section). Indeed, in these two cases, we have`T(ω) = 0 for every T >0 butCT(ω)>0 (i.e., we have observability) for T large enough. This is due to the fact that trapped rays are the weak limit of Gaussian beams that oscillateon both sides of the limit ray, spreading on one side and on the other a sufficient amount of energy so that indeed observability holds true. In full generality, having information on the way that semi-classical measures, supported on a grazing ray, can be approached by highfrequency wave packets such as Gaussian beams, is a difficult question. In the case of the sphere, symmetry arguments give the answer (see [8]). In the case of the torus, a much more involved analysis is required, based on second microlocalization arguments (see [3,5]).

Anyway, Corollary1can as well be applied for instance to any kind of checkerboard domainω on the two-dimensional torus, as soon as the measure ofω is large enough so that`T(ω)>1/2.

Since the case of checkerboards (in dimension two) is interesting and challenging, following a question by Nicolas Burq, in the next section we investigate the case of random checkerboards on the flat torus and we establish our second main result.

1.2 A probabilistic result for random checkerboards on the flat torus

In this section, we takeM =T2= IR2/Z2(flat torus) which is identified to the square [0,1]2, class of equivalence of IR2 under the identifications (x, y) ∼ (x+ 1, y) ∼ (x, y+ 1), inheriting of the Euclidean metric. Given any subset A of M, we denote by |A| the (two-dimensional) Lebesgue measure ofA.

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We consider a regular gridGn= (cij)16i,j6n in the square, like a checkerboard, made ofn×n closed squares:

[0,1]2=

n

[

i,j=1

cij with cij = [i−1, i]×[j−1, j] ∀(i, j)∈ {1, . . . , n}2.

Definingci0j0 in the same way for all (i0, j0)∈Z2, we identify the squareci0j0 to the squarecij of the above grid with (i, j)∈ {1, . . . , n}2such thati=i0 modnandj =j0 modn.

Construction of random checkerboards. Let ε ∈ [0,1] be arbitrary. Considering that all squares in the grid are initially white, we construct a random checkerboard by randomly darkening some squares in the checkerboard as follows: for every (i, j)∈ {1, . . . , n}2, we darken the square cij of the grid with a probability ε. All choices are assumed to be mutually independent. In other words, we make a selection of squares (that are paint in black) in the grid by considering n2 independent Bernoulli random variables denoted (Xij)16i,j6n, each of them with parameterε.

The total number of black squares follows therefore the binomial lawB(n2, ε).

We denote by ωnε the resulting (closed) subset of [0,1]2 that is the union of all black squares (see Figure1). Given any fixedT >0 andε∈(0,1], our objective is to understand how well the

Figure 1: Some examples of random checkerboards. The random subsetωnε ⊂[0,1]2 is the union of black squares.

random setωnε is able to capture all geodesic rays propagating inM '[0,1]2, in finite timeT. In other words, we want to study the random variable`Tnε). Of course, the random variable |ωεn| follows the law n12B(n2, ε) and thus its expectation is equal toε, and so, whenεis small,ωnε covers only a small area in [0,1]2. And yet, our second main result below shows that, fornlarge, almost all such random sets meet all geodesic rays within timeT.

Theorem 2. Given any T > 0 and any ε ∈ [0,1], the random variable `Tnε) converges in probability toε asn→+∞, i.e.,

n→+∞lim P |`Tεn)−ε|>δ

= 0 ∀δ >0.

Theorem 2 is proved in Section 3. As mentioned above, this issue has emerged following a question by Nicolas Burq. In [3,5], the authors also consider checkerboard domains, as above, but not in a random framework. As a consequence of their analysis, given anyT >0, anyε∈[0,1] and anyn∈IN fixed, if all geodesic rays of lengthT, either meet the interior ofωεn (i.e., the interior of some black square), or follow for some positive time one of the sides of a black square on the left and for some positive time one of the sides of a black square (possibly the same) on the right, thenCTnε)>0, i.e., the wave equation on the torusM =T2 is observable onωnε in time T.

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LetT >0 and letε∈(0,1] be arbitrary. According to Theorem2, forn large enough, almost every subsetωεn(constructed randomly as above) is such that`Tnε)>0. This implies that every geodesic ray, that is neither horizontal nor vertical, meets the interior of ωnε within time T, and that every horizontal or vertical geodesic ray meets the closed subsetωεn within timeT. In the latter case, moreover, by construction of the random setωnε, the probability that such grazing rays follow for some positive time one of the sides of a black square on the left and for some positive time one of the sides of a black square, converges to 1 asn→+∞.

All in all, combining Theorem 2, the result of [3, 5] and the above reasoning, we have the following consequence in terms of observability of the wave equation.

Corollary 2. Given any T >0 and ε∈(0,1], may they be arbitrarily small, the probability that the wave equation on the torusM =T2 is observable on ωεn in time T tends to1 asn→+∞.

In other words, observability in (any) finite time is almost surely true for largen, despite the fact that the measure ofωnε may be very small!

Note that, forε >1/2, almost sure observability follows from Corollary1 (indeed, the random sets constructed above are piecewiseC1 and thus Theorem 1 can be applied). But the result is more striking when|ωnε|is small.

Note also Theorem2provides an answer to an issue raised in [6], which we formulate in terms of an optimal shape design problem in the next corollary.

Corollary 3. For everyε∈[0,1], we have sup

|ω|6ε

`T(ω) =ε

where the supremum is taken over all possible measurable subsetsω ofM =T2 having a Lipschitz boundary.

Corollary3is proved in Section 3.4.

We finish this section by a comment on possible generalizations of Theorem 2. Some of the steps of its proof remain valid for any closed Riemannian manifold, like the fact that it suffices to prove the theorem forT small and thus, we expect that, to some extent, the result is purely local.

However, in some other steps we instrumentally use the fact that we are dealing with a regular checkerboard in the square. Extending the result to general manifolds, even in dimension two, is an open issue.

2 Additional comments and proof of Theorem 1

2.1 Comments on Theorem 1

Theorem1states that, given anyT >0 and any measurable subsetω ofM, we have

`T(ω)61

2 `T(˚ω) + 1 .

under the two following sufficient assumptions:) the metric g is C2; (ii) ω is an embedded C1 submanifold ofM with boundary ifn>3 and is piecewiseC1 ifn= 2.

Remark 1. Assumption (ii) may be weakened as follows:

• IfM is of dimension 2, it suffices to assume thatωis piecewiseC1. More precisely, we assume thatω is a C1 stratified submanifold of M (in the sense of Whitney).

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• In any dimension, the following much more general assumption is sufficient: given any grazing rayγ, for almost everyt∈[0, T] such thatγ(t)∈∂ω, the subdifferential atγ(t) of∂ω∩γ(·) is a singleton. This is the case under the (much stronger) assumption thatω be geodesically convex.

Comments. It is interesting to note that the assumptions made in Theorem 1 are essentially sharp. Remarks are in order.

• The inequality (4) gives a quantitative measure of the discrepancy that can happen for `T when we take the closure of a measurable subsetωor, conversely, when we pass to the interior (this is the sense of Corollary1). The inequality is sharp, as shown by the example already discussed above: take M = S2 and ω the open Northern hemisphere; then `T(ω) = 0 for everyT >0 and`(ω) = 1/2 for T = 2π. Hence, here, (4) is an equality.

• As a variant, takeωwhich is the union of the open Northern hemisphere and of a Southern spherical cap, i.e., a portion of the open Southern hemisphere limited by a given latitude

−ε <0. Then we have as well `T(ω) = 0 for everyT >0 and`(ω) = 1/2 for T = 2π.

• Note that, takingε = 0 in the previous example (i.e., ω is the unit sphere M =S2 minus the equator), we have`T(ω) = 0 and`T(ω) = 1 for everyT >0 and thus (4) fails. But here, ω is not an embeddedC1 submanifold ofM with boundary: Assumption (ii) (which implies local separation between ˚ω andM \ω) is not satisfied. More generally, the result does not apply to any subsetω that is M minus a countable number of rays. This is as well the case when one considers any subsetωthat is dense and of empty interior (one has`T(˚ω) = 0 and

`T(ω) = 1 for every T >0). This shows that the discrepancy 1/2 is only valid under some regularity assumptions onω.

• There is no discrepancy in the absence of geodesic rays grazingω: if there is no geodesic ray having a contact of infinite order with∂ω=ω\˚ω then`T(ω) =`T(˚ω).

• The result fails in general if∂ω is piecewiseC1 only, on a manifoldM is of dimensionn>3.

Here is a counterexample.

Let γ be a geodesic ray. If T > 0 is small enough, it has no conjugate point. In a local chart of coordinates, we have γ(t) = (t,0, . . . ,0) (see the proof of Theorem1). Now, using this local chart we define a subset ω of M as follows: the section of ∂ω with the vertical hyperplaneγ(·) is locally equal to this entire hyperplane minus a cone of vertexγ(t) with small angle 2πε > 0, less than π/4 for instance (see Figure 2). Now, we assume that, as t >0 increases, these sections rotate with such a speed that, along [0, T], the entire vertical hyperplane is scanned by the section withω. If the speed of rotation is exactlyT /2πthen it can be proved that`T(˚ω) = 0 and`T(ω) = 1−2ε.

This example shows that Assumption (ii), or its generalization given in Remark1, cannot be weakened too much. The idea here is to consider a subsetωsuch that the section of∂ωwith the vertical hyperplane γ(·) has locally the shape of the hypograph of an absolute value, which is rotating alongγ(·).

Similar examples can as well be designed with checkerboard-shaped domainsω, thus under- lining that in [3,5] it was important to consider checkerboards in dimension 2.

• Surprisingly, the result is wrong if the metricgis notC2. A counterexample is the following.

LetM be a pill-shaped two-dimensional manifold given by the union of a cylinder of finite length, at the extremities of which we glue two hemispheres (domain also obtained by rotating

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∂ωγ(·) γ(0)

γ(t)

γ(T)

∂ωγ(·) ∂ωγ(·)

Figure 2: Locally aroundγ(t),∂ω∩γ(t) is the complement of the hatched area.

a 2D stadium in IR3around its longest symmetry axis; or, take the unit sphere in IR3, cut it at the equator, separate the two hemispheres and glue them with, inbetween, a cylinder of arbitrary length), and endow it with the induced Euclidean metric (see Figure3). Then the

γ(·) ω

Figure 3: M is pill-shaped andω is the complement of the hatched area.

metric is notC2 at the gluing circles. Now, takeω defined as the union of the open cylinder with two open spherical caps (i.e., the union of the two hemispheres of which we remove latitudes between 0 and some ε >0). Then `T(ω) = 0 for everyT >0, because ω does not contain the rays consisting of the circles at the extremities of the cylinder. In contrast,`T(ω) may be arbitrarily close to 1 asT is large enough andε is small enough, and thus (4) fails.

This is because any ray ofM spending a timeπ in M\ω spends then much time over the cylinder.

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This shows that Assumption (i) is sharp. In the above example, the metric is onlyC1,1. The example above is rather counterintuitive. The assumption of a C2 metric implies in some sense a global result on geodesic rays.

Our proof, given in Section 2.2 hereafter, uses only elementary arguments of Riemannian ge- ometry. It essentially relies on Lemma2, in which we establish that, given a grazing ray (i.e., a ray propagating in ∂ω), thanks to our assumption on ω we can always construct neighbor rays, one of which being insideω and the other being outside ofω for all times.

2.2 Proof of Theorem 1

Without loss of generality, we takeω⊂Mopen. We will use several well known facts of Riemannian geometry, for which we refer to [2].

Lemma 1. There exists γ ∈ Γ such that `T(ω) = T1RT

0 χω(γ(t))dt, i.e., the infimum in the definition (1)of `T(ω)is reached.

Proof. Let (γk)k∈IN be a sequence of geodesic rays such that T1 RT

0 χωk(t))dt → `T(ω). By compactness of geodesics,γk(·) converges uniformly to some geodesic rayγ(·) on [0, T].

Let t ∈[0, T] be arbitrary. If γ(t)∈ ω then for k large enough we have γk(t) ∈ω, and thus 1 =χω(γ(t))6χωk(t)) = 1. If γ(t)∈ M \ω then 0 = χω(γ(t))6χωk(t)) for any k. In all cases, we have obtained the inequality

χω(γ(t))6lim inf

k→+∞χωk(t)) for everyt∈[0, T].

By the Fatou lemma, we infer that

`T(ω)6 1 T

Z T 0

χω(γ(t))dt6 1 T

Z T 0

lim inf

k→+∞χωk(t))dt6lim inf

k→+∞

1 T

Z T 0

χωk(t))dt=`T(ω).

The lemma follows.

If the rayγgiven by Lemma1is not grazing, i.e., ifRT

0 χ∂ω(γ(t))dt= 0, thenRT

0 χω(γ(t))dt= RT

0 χω(γ(t))dt and thus`T(ω)6 T1

RT

0 χω(γ(t))dt6`T(ω) and hence `T(ω) =`T(ω). So in this case there is nothing to prove.

In what follows we assume that the rayγgiven by Lemma1is grazing, i.e.,RT

0 χ∂ω(γ(t))dt >0.

Assume thatγ(t) = π◦ϕt(x0, ξ0) with x0 ∈ M and ξ0 ∈ Sx

0M. Here, Sx

0M denotes the unit cotangent bundle atx0(i.e.,kξ0kg?= 1 whereg is the cometric),ϕtis the geodesic flow onSM andπ:SM →M is the canonical projection.

Lemma 2. There exists a continuous path of points s7→xs ∈M, passing through x0 at s= 0, such that, settingγs(t) =π◦ϕt(xs, ξ0), we have

s→0lim(χωs(t)) +χω−s(t))) = 1 (5) for almost everyt∈[0, T] such thatγ(t)∈∂ω.

Proof. To prove this fact, we assume that, in a local chart, γ(t) = (t,0, . . . ,0). This is true at least in a neighborhood of x0 = γ(0) = 0, and this holds true along γ(·) as long as there is no conjugate point. We also assume that, in this chart, any other geodesic ray starting at

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(0, x02, . . . , x0n) in a neighborhood ofγ(0) = (0, . . . ,0) is given by (t, x02, . . . , x0n) (projection onto M of the extremal field). This classical construction of the so-called extremal field can actually be done on any subinterval of [0, T] along which there is no conjugate point. Note that the set of conjugate times along [0, T] is of Lebesgue measure zero1. Let us search an appropriate (n−1)- tuple (x02, . . . , x0n)∈IRn−1\ {0}such that the family of points xs= (0, sx02, . . . , sx0n),s∈(−1,1), gives (5). Note that the geodesic ray starting at xs isγs(t) = (t, sx02, . . . , sx0n) in the local chart.

In what follows, we set N = ∂ω = ω \˚ω = ω \ω (ω is open). By assumption, ω is an embeddedC1 submanifold ofM with boundary and one has dimN = dimM−1. By assumption, in a neighborhoodU of any point of N, the set N∩U is a codimension-one hypersurface of M, written asF = 0 withF :U →IR of classC1, which is separatingω and M\ω in the sense that ω∩U ={F <0},N ={F = 0}and M\ω={F >0}.

It suffices to prove that, for almost every timetat whichγ0(t) =γ(t)∈N and ˙γ(t)∈Tγ(t)N, the pointsγs(t) andγ−s(t) are on different sides with respect to the (locally) separating manifold N forssmall enough.

This is obvious whenγ is transverse toN. We set Ω ={t∈[0, T] | γ(t)∈N, γ(t)˙ ∈Tγ(t)N}.

It is a closed subset of [0, T]. Lett∈Ω. In the local chart the tangent spaceTγ(t)Nis an hyperplane of IRncontaining the line IR(1,0, . . . ,0). Its projection onto{0}×IRn−1(the hyperplane orthogonal to the lineγ(·)) is an hyperplane of {0} ×IRn−1, of normal vector (0, v(t)) with v(t)∈ IRn−1 of Euclidean norm 1. Since only the direction ofv(t) is important, we assume thatv(t)∈Pn−2(IR), the projective space.

We claim that:

There existsV ∈Pn−2(IR)such that hV, v(t)i 6= 0 for almost everyt∈Ω.

With this result, settingV = (x02, . . . , x0n), the pointsxs defined above give the lemma.

Let us now prove the claim. We define A ={(t, V) ∈ Ω×Pn−2(IR) | hV, v(t)i = 0}. By definition, given (t, V)∈Ω×Pn−2(IR) we haveχA(t, V) = 1 whenV ∈v(t). Sincev(t)∩Pn−2(IR) is of codimension one inPn−2(IR), we haveR

Pn−2(IR)χA(t, V)dHn−2= 0 for everyt∈Ω, where we have endowedPn−2(IR) with the Hausdorff measureHn−2. Therefore, by the Fubini theorem,

0 = Z

Z

Pn−2(IR)

χA(t, V)dHn−2dt= Z

Pn−2(IR)

Z

χA(t, V)dt dHn−2 and thus R

χA(t, V)dt = 0 for almost every V ∈ Pn−2(IR). Fixing such a V, it follows that χA(t, V) = 0 for almost everyt∈Ω, and the claim is proved.

In view of proving Remark1, note that the argument above still works in dimension 2 withω piecewiseC1 (but not in dimension greater than or equal to 3: see the counterexample given in Section 1). In more general, in any dimension, the argument above still works if ω is such that, for almost every timet, the subdifferential atγ(t) of∂ω∩γ(·) is a singleton.

At this step, we have embedded the rayγ given by Lemma 1 into a family of rays γs which enjoy a kind of transversality property with respect toN =∂ω. Let us consider the partition

[0, T] =A1∪A2∪A3

into three disjoint measurable sets, with

A1={t∈[0, T] | γ(t)∈ω}, A2={t∈[0, T] | γ(t)∈M\ω}, A3={t∈[0, T] | γ(t)∈∂ω}.

1This is a general fact in Riemannian geometry. Indeed, a conjugate time is a time at which a non-zero Jacobi field vanishes. Since Jacobi fields are solutions of a second-order ordinary differential equation, such times must be isolated, for otherwise the Jacobi field would vanish at the second order and thus would be identically zero.

(11)

Sinceγs(·) converges uniformly toγ(·) ass→0 and sinceω andM\ω are open, we have:

• lims→0ωs(t)) +χω−s(t))) = 2 for everyt∈A1;

• lims→0ωs(t)) +χω−s(t))) = 0 for everyt∈A2;

• lims→0ωs(t)) +χω−s(t))) = 1 for almost everyt∈A3 (by Lemma2).

By the Lebesgue dominated convergence theorem, we infer that

s→0lim Z T

0

ωs(t)) +χω−s(t)))dt= 2|A1|+|A3|.

Now, on the one part, by the first step we have T1|A1| = T1RT

0 χω(γ(t)) =`T(ω). On the other part, sinceA1 andA3 are disjoint we have T1(|A1|+|A3|)61. Hence

s→0lim 1 T

Z T 0

ωs(t)) +χω−s(t)))dt6`T(ω) + 1.

Since `T(ω) 6 T1

RT

0 χω±s(t))dt for every s by definition, we infer that 2`T(ω) 6 `T(ω) + 1.

Theorem1is proved.

3 Proof of Theorem 2

Theorem2states that, given anyT >0 and anyε∈[0,1], the random variable`Tnε) converges in probability toεasn→+∞, i.e.,

n→+∞lim P |`Tεn)−ε|>δ

= 0 ∀δ >0.

This section is organized as follows. We make a preliminary remark in Section3.1. In Section3.2, we give the successive steps of the proof, involving intermediate lemmas that are proved. One of the main ingredients of the proof of Theorem2 is a large deviation property which is established in Section3.3. In Section3.4, we also provide a proof of Corollary3.

3.1 Preliminaries

For any geodesic rayγ∈Γ and any measurable subsetω⊂M, we set mTγ(ω) = 1

T Z T

0

χω(γ(t))dt so that we have

`T(ω) = inf

γ∈ΓmTγ(ω).

Lemma 3. Given any measurable subset ω ⊂ M, the mapping Γ 3 γ 7→ mTγ(ω) is upper semi- continuous.

Given any ω that is a union of squares from the grid Gn with n fixed, the mapping Γ 3γ 7→

mTγ(ω)is continuous at each γ which

• is not horizontal or vertical;

• is horizontal and vertical but meets no corner.

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Proof. Upper semi-continuity in general is obvious. Let us prove the second part of the lemma.

Let (γk)k∈INbe a sequence of geodesic rays converging toγ∈Γ. We assume that either thatγis not horizontal or is not vertical, or thatγ is horizontal or vertical and meets no corner. Writing

χω=

n

X

i,j=1

τ(i, j)χcij

whereτ(i, j) = 1 ifcij ⊂ω andτ(i, j) = 0 otherwise, and noting that mTγ

k(ω) =

n

X

i,j=1

τ(i, j)mTγ

k(cij)

we have to prove thatmTγk(cij) converges tomTγ(cij) ask→+∞. This follows from the dominated convergence theorem and from the fact that (χcijk))k∈INconverges almost everywhere toχcij(γ).

The latter claim can be shown by distinguishing between two cases: ifγ(t)∈˚cij then fork large enough we haveγk(t)∈˚cij andχcijk(t)) =χcij(γ(t)) = 1 for suchk. The same conclusion occurs ifγ(t)∈/ cij. Since the set of pointst such thatγ(t)∈∂cij is finite (the speed of the geodesic ray is equal to 1), the lemma follows.

Since γ 7→ mTγ(ω) is only upper semi-continuous but not continuous, it will actually be con- venient to slightly modify its definition when γ is horizontal (or vertical) and meets a corner by setting

mTγ(ω) = inf

0k)k∈IN

lim inf

k→+∞mTγ0

k(ω) (6)

where the infimum is taken on the set of sequences (γ0k)k∈INof horizontal (or vertical) geodesic rays converging toγ. With this modification, the mapping γ 7→mTγ(ω) is now lower semi-continuous and thus is continuous, and we still do have`T(ω) = infγ∈ΓmTγ(ω) because the value ofmTγ(ω) at possible points of discontinuity is changed into a value of the closure of the range of the mapping.

In particular, since the set of geodesic rays of lengthT is compact, the infimum in the definition of`T(ω) is reached for someγ∈Γ, i.e., we have`T(ω) =mTγ(ω).

3.2 Proof of Theorem 2

For every (i, j)∈ {1, . . . , n}2, letXij be the random variable equal to 1 whenevercij ⊂ωnε and 0 otherwise. Recall that, assuming that all square cells of the grid are initially white, when ranging over the grid, for each cell we randomly darken the cell with a probabilityε(Bernoulli law) and in this case we setXij = 1; otherwise we letXij= 0. By construction, the random laws (Xij)16i,j6n are independent and identically distributed (i.i.d.), of expectationεand of varianceε(1−ε).

Given any fixed geodesic rayγ∈Γ, we denote bytij(γ) the time spent byγin the square cell cij. We havePn

i,j=1tij(γ) =T. It may happen thatγcrosses several times (at mostT+ 1 times) the square cellcij, and in this case tij(γ) is the sum of several passage times; since the maximal time spent incij by a ray in one passage is

2

n , we havetij(γ)6(T+ 1)

2

n . Summing up, we have

∀(i, j)∈ {1, . . . , n}2 tij(γ)

T 6T+ 1 T

√2

n and

n

X

i,j=1

tij(γ)

T = 1 (7)

and we also note that

mTγεn) = 1 T

Z T 0

χωεn(γ(t))dt=

n

X

i,j=1

tij(γ)

T Xij. (8)

(13)

In particular, the random variablemTγnε) is a weighted sum of independent Bernoulli laws, and thus

EmTγnε) =ε (9)

and

VarmTγnε) =

n

X

i,j=1

tij(γ) T

2

ε(1−ε)6T+ 1 T

√2ε(1−ε)

n (10)

where we have used (7) and the fact that 06tijT(γ) 61.

To prove Theorem2, we have to prove that, given anyT >0 and anyε∈[0,1], for everyδ >0 we have

(i) lim

n→+∞P `Tεn)6ε+δ

= 1, (ii) lim

n→+∞P `Tεn)>ε−δ

= 1, or equivalently, lim

n→+∞P `Tnε)6ε−δ

= 0.

Proof of (i). Using (8) and by definition of`T which is an infimum over all rays, we have

`Tnε)6mTγnε) =

n

X

i,j=1

tij(γ)

T Xij. (11)

Applying the Bienaym´e-Tchebychev inequality to the random variablemTγεn), we have P |mTγεn)−EmTγnε)|>δ

6VarmTγnε) δ2

and thus, using (9) and (10),

n→+∞lim P mTγnε)6ε+δ

= 1. (12)

Finally,(i)follows from (11) and (12).

Proof of(ii). Establishing (ii)is much more difficult. We proceed in several steps, by proving the following successive lemmas that are in order.

Lemma 4. If (ii) is true for everyT ∈(0,1), then it is true for everyT >0.

Thanks to Lemma4 (proved further), we now assume that 0< T <1.

We introduce some notations. Let Γ0, Γ1and Γ2be the sets of geodesic rays ofM =T2meeting respectively zero, at least one and at least two corners of the gridGn (by definition, a corner is a point (i/n, j/n) in [0,1]2, for some (i, j)∈ {0, . . . , n}2). Given anyω ⊂M =T2that is the union of disjoint closed square cells ofGn, we have

`T(ω) = inf

γ∈Γ0∪Γ1mTγ(ω).

We also define

`TΓ1(ω) = inf

γ∈Γ1mTγ(ω), `TΓ2(ω) = inf

γ∈Γ2mTγ(ω).

Of course, we have

`T(ω)6`TΓ1(ω)6`TΓ2(ω).

(14)

Lemma 5. We have |`Tnε)−`TΓ1nε)|= O(1/n)asn→+∞.

Lemma 6. We have |`TΓ1εn)−`TΓ2εn)|= O(1/n)asn→+∞.

Lemma 7. We have lim

n→+∞P `TΓ2εn)6ε−δ

= 0.

Finally,(ii)follows from the above lemmas, that are proved hereafter.

3.2.1 Proof of Lemma 4

LetT >1 and letm∈IN be such thatT0= mT ∈(0,1). Let us prove that

`Tnε)>`T0εn). (13) Given anyρ >0, letγbe a geodesic ray such that

`Tnε) +ρ> 1 T

Z T 0

χωnε(γ(t))dt.

Settingγk(·) =γ(kT0+·) for everyk∈ {0, . . . , m−1}, we have

`Tnε) +ρ> 1 T

Z T 0

χωn

ε(γ(t))dt= 1 T

m−1

X

k=0

Z T0 0

χωn

εk(t))dt> 1 T

m−1

X

k=0

T0`T0εn) =`T0εn) Lettingρtend to 0, we obtain (13).

Since 0< T0 <1,(ii)is true for this final timeT0, i.e.,

n→+∞lim P

`T0nε)>ε−δ

= 1.

Therefore, using (13), we obtain (ii)for the final timeT.

3.2.2 Proof of Lemma 5

Letωbe a subset of [0,1]2that is a union of square cells of the gridGn. Recall that`T(ω)6`TΓ1(ω).

Now, let us prove that`TΓ1(ω)6`T(ω) +Cn whereC >0 does neither depend onnnor onω(which is a stronger statement, implying the lemma). To this aim, letγ∈Γ be such that

mTγ(ω)< `T(ω) + 1

nT. (14)

Ifγ∈Γ1then we are done. Hence, in what follows we assume thatγ∈Γ0. Moreover, sinceγmeets no corner, rotating slightlyγ if necessary, by continuity ofmTγ (see the modified definition (6) of mTγ in Section3.1) we can assume thatγ is neither vertical nor horizontal and still satisfies (14).

Letnbe a unit vector orthogonal toγ0(0). For everys∈IR, we denote byTsnthe translation of vectorsnand we define the translated geodesic rayγs=Tsn◦γ. By continuity,γsmeets the same square cells asγ if|s|is small enough. LetI(γ) denote the subset of all indices (i, j)∈ {1, . . . , n}2 such thatγcrosses the cell squarescij. If|s|is small enough then

mTγs(ω) = X

(i,j)∈I(γ)

tijs) T Xij

where tijs) is the time spent by γs in cij. Now, note that for given indices i and j such that γ(0), γ(T)∈/˚cij, an easy geometric argument shows thats7→tijs) is affine and nonconstant with

(15)

respect tos. Hence, denoting byI0(γ) the set of (i, j)∈I(γ) such thatγ(0), γ(T)∈/˚cij (note that

#I(γ)−26#I0(γ)6#I(γ)), we define Mγs(ω) = 1

T X

(i,j)∈I0(γ)

tijs)Xij.

We have

Mγs(ω)6mTγs(ω)6Mγs(ω) + X

(i,j)∈I(γ)\I0(γ)

tijs)

T Xij 6Mγs(ω) +2√ 2 nT .

Since tijs) is an affine function for (i, j) ∈ I0(γ), so is s 7→ Mγs(ω). Replacing s by −s if necessary, we infer the existence ofs0 >0 such that the mapping s 7→ Mγs(ω) is decreasing on (0, s0) and moreover γs0 ∈Γ1. Sinceγ is neither vertical nor horizontal, so isγs0. We then infer that

mTγs

0(ω)6Mγs0(ω) +2√ 2 nT = lim

s→s0

Mγs(ω) +2√ 2

nT 6mTγ(ω) +2√ 2 nT . Using (14), we obtain

`TΓ1(ω)6mTγs

0(ω)< `T(ω) +2√ 2 + 1 nT . The conclusion follows.

3.2.3 Proof of Lemma 6

Letωbe a subset of [0,1]2that is a union of square cells of the gridGn. Recall that`TΓ1(ω)6`TΓ2(ω).

Now, let us prove that`TΓ2(ω)6`TΓ1(ω) +Cn whereC >0 does neither depend onnnor onω(this is a stronger statement, implying the lemma). To this aim, let γ ∈ Γ12. It suffices to find γ0∈Γ2 such that

mTγ0(ω)6mTγ(ω) +C

n. (15)

By definition, the geodesic rayγ meets exactly one corner of the grid, that we denote by O. We denote byRθthe rotation centered atOwith angleθand we define the geodesic rayγθ=rθ◦γ. As in the proof of Lemma5, letI0(γ) be the set of all indices (i, j)∈ {1, . . . , n}2 such thatγθ crosses cij andγ(0), γ(T)∈/˚cij. This condition does not depend onθ ∈J = (−θ, θ+) whereθ, θ+ >0 are such thatγθ∈/Γ2for everyθ∈J. As before, the reason why we require thatγ(0), γ(T)∈/˚cij is technical: it allows us to make easier the computation of the timetijθ) spent byγθincij. As in the proof of Lemma5, defining

Mγθ(ω) = 1 T

X

(i,j)∈I0(γ)

tijθ)Xij (16)

where we recall thatXij is equal to 1 ifcij ⊂ω and 0 otherwise, we have Mγθ(ω)6mTγ

θ(ω)6Mγθ(ω) +2√ 2

nT . (17)

We can assume that the real numbers θ± above are such that J is “maximal”, meaning that γθ± ∈Γ2. Now, we define the functionf onJ by

f(θ) =Mγθ(ω)

(16)

and we extend it by continuity on ¯J. It remains to prove that that min(f(−θ), f(θ+))6min

J f+C

n (18)

whereC does neither depend onn nor onγ. Indeed, assuming that (18) is satisfied, let us finish the proof of Lemma6. Without loss of generality, we assume thatf(−θ)6f(θ+). Note that the modified definition (6) ofmTγ (in Section3.1), (16) and (17) imply thatmTγ

θ(ω)6f(−θ) +2

2 nT . This inequality is actually an equality unlessγθ is horizontal or vertical. Therefore

mTγ

θ(ω)6min

J f+C n +2√

2 nT .

Since minJf 6f(0) =Mγ(ω)6mTγ(ω), we obtain mTγθ

(ω)6mTγ(ω) + Cn0 with C0 =C+2

2 T . Sinceγθ ∈Γ2, the inequality (15) is satisfied and therefore Lemma6 is proved.

It remains to prove (18). We first computef by providing an explicit expression oftijθ). We claim that, if (i, j)∈I, then there exists (aij, bij)∈ {−n, . . . , n}2 such that

tijθ) = aij

nsin(θ0+θ)+ bij

ncos(θ0+θ) (19)

whereθ0is the angle between the horizontal axis and the geodesic rayγ. Without loss of generality we assume thatθ0∈[0, π/2]. Note thatθ0∈(0, π/2), because otherwise the two conditionsγ∈Γ1

andθ0 ∈ {0, π/2} (i.e., γ is horizontal or vertical) would imply that γ ∈Γ2, which is false. This also implies thatJ ⊂(0, π/2). Using the notations of Figure4 where the corner O is assimilated

Figure 4: Particular geodesics issued fromO and meeting the square with bold boundary.

to the origin, and focusing on the squarecij whose coordinates are (α/n, β/n), (α/n+ 1/n, β/n),

(17)

(α/n+ 1/n, β/n+ 1/n) and (α/n, β/n+ 1/n), an elementary geometric reasoning gives

tijθ) =





α+1

ncosθnsinβ θ ifθ0+θ∈[θ1, θ2]

1

nsinθ ifθ0+θ∈[θ2, θ3]

β+1

nsinθncosα θ ifθ0+θ∈[θ3, θ4]

where 0< θ1< θ2< θ3< θ4< π/2 are the angles between the horizontal axis and the particular geodesics passing at the corners ofcij (see Figure4). Note that this formula is still satisfied when θ1= 0 orθ4=π/2. In such cases, eitherαor β vanishes. The claim (19) is then proved.

Note that the rotation Rθ can be applied until the geodesic rayγθ meets a new corner. It is then easy to see that

θ+, θ6 C

n (20)

whereCdoes not depend onnandγ.

According to (19), we have

Mγθ(ω) = X1

sin(θ0+θ)+ X2

cos(θ0+θ), (21)

with

X1= 1 n

X

(i,j)∈I0(γ)

aijMγθ(cij) and X2= 1 n

X

(i,j)∈I0(γ)

bijMγθ(cij), forθ∈J. Moreover,

|X1|6#I0(γ)62nT and |X2|6#I0(γ)62nT because max(|aij|,|bij|)6nT andMγθ(cij)6

2 nT.

We are now in a position to establish (18). It is is obvious if minJf is reached at θor θ+. Let us then assume thatf reaches its minimum; without loss of generality, we assume that the minimum is reached atθ= 0. In particular,

d dθ

θ=0

Mγθ(ω) = 0

which, according to (21), givessincos(θ200))X1=cossin(θ200))X2. From (21), we also havef(0) = sinθX1

0+cosX2θ

0

and thusX1=f(0) sin30) andX2=f(0) cos30), so that f(θ) =Mγθ(ω) =f(0)

sin30)

sin(θ0+θ)+ cos30) cos(θ0+θ)

.

Now,

f(θ)−f(0) =f(0)

sin3θ0

1

sin(θ0+θ)− 1 sinθ0

+ cos3θ0

1

cos(θ0+θ)− 1 cosθ0

.

This computation shows that θ0+ 6=π/2 or θ0−θ 6= 0. Otherwise, since θ0 ∈ {0, π/2},/ f would be not bounded onJ whenθ tends toθ0+ or θ0−θ, which is wrong by definition of f. This implies that

arcsin 1

nT

0−θ< θ0+6 π

2 −arcsin 1

nT

. (22)

(18)

Indeed, set γ = γθ0−θ. We know that γ([0, T]) contains at least two corners and is not horizontal. This means that its second coordinate (inR2) increases at least of 1/nbetween 0 and T and henceTsin(θ0−θ)>1/n. The same argument allows to prove the last inequality of (22).

From now on, we letntend to +∞. In particular,θ0 depends onn. Let us prove that f(θ)−min

J f =f(θ)−f(0)6 C

n (23)

A similar argument will show thatf(θ+)−minJf 6 Cn. These two estimates imply (18) and hence complete the proof of Lemma6. It thus remains to prove (23). By the mean value theorem, there exists ˜θ∈(0, θ) such that

1

sin(θ0−θ)− 1

sinθ0 = −cos(θ0−θ)˜ sin20−θ)˜ θ. Since 06θ˜6θ, using that|sin(θ0) cos(θ0−θ)|˜ 61, we get from (20) that

sin30)

1

sin(θ0−θ)− 1 sinθ0

6 C

n

sin(θ0) sin(θ0−θ)˜

2 .

Now, by the mean value theorem, sin(θ0)

sin(θ0−θ)˜ = 1 + sin(θ0)−sin(θ0−θ)˜

sin(θ0−θ)˜ 61 + O(1/n) sin(θ0−θ)˜ .

Using (22), there existsC0 >0 which does neither depend onγnor onnsuch that O(1/n)

sin(θ0θ)˜ = O(1).

This implies that

sin30)

1

sin(θ0−θ)˜ − 1 sinθ0

= O 1

n

.

Following the same reasoning, we prove that cos30)

1

cos(θ0−θ)˜ − 1 cosθ0

= O 1

n

.

Actually, this last estimate is much easier since|cos(θ0)|6Ccos(θ0−θ). This leads to (23) and concludes the proof of Lemma6.

3.2.4 Proof of Lemma 7

Let (i1, j1),(i2, j2)∈ {1, . . . , n}2 be such that the two square cellsci1j1 andci2j2 ofGnare disjoint and letx1andx2 be two corners of the gridGn. Any geodesic ray of lengthT such that

γ(0)∈ci1j1, γ(T)∈ci2j2 and xi∈γ([0, T]), i= 1,2, will be denoted byγci1j1,ci2j2,x1,x2.

Let ˆΓ be the set of all such geodesic raysγci1j1,ci2j2,x1,x2, as (i1, j1), (i2, j2), x1 and x2 vary.

By construction, we have #ˆΓ4= O(n8).

Let ˆγ∈ˆΓ. By the large deviation result established in Proposition1 in Section3.3(it can be applied thanks to (7)), we have

P(mγˆnε)6ε−δ)6Cε,δe

2 c

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